Exercise 11.1
If a line makes angles 90°, 135°, 45° with the x, y and z-axes respectively, find its direction cosines.
Answer
Direction cosines are l = cos 90°, m = cos 135°, n = cos 45°.
l = 0 → m = −1/√2 → n = 1/√2
Direction cosines: 0, −1/√2, 1/√2.
Find the direction cosines of a line which makes equal angles with the coordinate axes.
Answer
Let each angle be α, so l = m = n = cos α.
l² + m² + n² = 1
⇒ 3cos²α = 1
⇒ cos α = ±1/√3
Direction cosines: (1/√3, 1/√3, 1/√3) or (−1/√3, −1/√3, −1/√3).
If a line has the direction ratios −18, 12, −4, then what are its direction cosines?
Answer
√[(−18)² + 12² + (−4)²] = √(324 + 144 + 16) = √484 = 22
l = −18/22 = −9/11, m = 12/22 = 6/11, n = −4/22 = −2/11
Direction cosines: −9/11, 6/11, −2/11 (or their negatives, 9/11, −6/11, 2/11, for the opposite sense).
Show that the points (2, 3, 4), (−1, −2, 1), (5, 8, 7) are collinear.
Answer
Let A(2, 3, 4), B(−1, −2, 1), C(5, 8, 7).
Direction ratios of AB: −1 − 2, −2 − 3, 1 − 4 = −3, −5, −3
Direction ratios of BC: 5 + 1, 8 + 2, 7 − 1 = 6, 10, 6
Since 6/(−3) = 10/(−5) = 6/(−3) = −2, the direction ratios are proportional, so AB ∥ BC. B is common to both, hence A, B, C are collinear.
Find the direction cosines of the sides of the triangle whose vertices are (3, 5, −4), (−1, 1, 2) and (−5, −5, −2).
Answer
Let A(3, 5, −4), B(−1, 1, 2), C(−5, −5, −2).
Side AB: direction ratios −4, −4, 6
⇒ AB = √(16 + 16 + 36) = √68 = 2√17
⇒ direction cosines −2/√17, −2/√17, 3/√17.
Side BC: direction ratios −4, −6, −4
⇒ BC = √(16 + 36 + 16) = √68 = 2√17
⇒ direction cosines −2/√17, −3/√17, −2/√17.
Side CA: direction ratios 8, 10, −2
⇒ CA = √(64 + 100 + 4) = √168 = 2√42
⇒ direction cosines 4/√42, 5/√42, −1/√42.
Exercise 11.2
Show that the three lines with direction cosines 12/13, −3/13, −4/13; 4/13, 12/13, 3/13; 3/13, −4/13, 12/13 are mutually perpendicular.
Answer
Two lines are perpendicular if l₁l₂ + m₁m₂ + n₁n₂ = 0.
Lines 1 and 2: (12×4 − 3×12 − 4×3)/169 = (48 − 36 − 12)/169 = 0
Lines 2 and 3: (4×3 − 12×4 + 3×12)/169 = (12 − 48 + 36)/169 = 0
Lines 1 and 3: (12×3 + 3×4 − 4×12)/169 = (36 + 12 − 48)/169 = 0
Hence the three lines are mutually perpendicular.
Show that the line through the points (1, −1, 2), (3, 4, −2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).
Answer
Direction ratios of first line: 3 − 1, 4 + 1, −2 − 2 = 2, 5, −4.
Direction ratios of second line: 3 − 0, 5 − 3, 6 − 2 = 3, 2, 4.
a₁a₂ + b₁b₂ + c₁c₂ = 2(3) + 5(2) + (−4)(4) = 6 + 10 − 16 = 0. Hence the lines are perpendicular.
Show that the line through the points (4, 7, 8), (2, 3, 4) is parallel to the line through the points (−1, −2, 1), (1, 2, 5).
Answer
Direction ratios of first line: 2 − 4, 3 − 7, 4 − 8 = −2, −4, −4.
Direction ratios of second line: 1 + 1, 2 + 2, 5 − 1 = 2, 4, 4.
−2/2 = −4/4 = −4/4 = −1, so the direction ratios are proportional and the lines are parallel.
Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector 3î + 2ĵ − 2k̂.
Answer
a = î + 2ĵ + 3k̂, b = 3î + 2ĵ − 2k̂. Using r = a + λb:
r = (î + 2ĵ + 3k̂) + λ(3î + 2ĵ − 2k̂)
Cartesian form: (x − 1)/3 = (y − 2)/2 = (z − 3)/(−2).
Find the equation of the line in vector and in cartesian form that passes through the point with position vector 2î − ĵ + 4k̂ and is in the direction î + 2ĵ − k̂.
Answer
Vector form: r = (2î − ĵ + 4k̂) + λ(î + 2ĵ − k̂)
Cartesian form: point (2, −1, 4), direction ratios 1, 2, −1
⇒ (x − 2)/1 = (y + 1)/2 = (z − 4)/(−1).
Find the cartesian equation of the line which passes through the point (−2, 4, −5) and parallel to the line given by (x + 3)/3 = (y − 4)/5 = (z + 8)/6.
Answer
The required line has the same direction ratios 3, 5, 6.
(x + 2)/3 = (y − 4)/5 = (z + 5)/6
The cartesian equation of a line is (x − 5)/3 = (y + 4)/7 = (z − 6)/2. Write its vector form.
Answer
The line passes through (5, −4, 6) and has direction ratios 3, 7, 2.
r = (5î − 4ĵ + 6k̂) + λ(3î + 7ĵ + 2k̂)
Q8. Find the angle between the following pairs of lines
(i) r = 2î − 5ĵ + k̂ + λ(3î + 2ĵ + 6k̂) and r = 7î − 6k̂ + μ(î + 2ĵ + 2k̂)
(ii) r = 3î + ĵ − 2k̂ + λ(î − ĵ − 2k̂) and r = 2î − ĵ − 56k̂ + μ(3î − 5ĵ − 4k̂)
Ans. cos θ = |b₁·b₂| / (|b₁||b₂|)
(i) b₁ = 3î + 2ĵ + 6k̂, b₂ = î + 2ĵ + 2k̂ → b₁·b₂ = 3 + 4 + 12 = 19 → |b₁| = 7, |b₂| = 3 → cos θ = 19/21 → θ = cos⁻¹(19/21).
(ii) b₁ = î − ĵ − 2k̂, b₂ = 3î − 5ĵ − 4k̂ → b₁·b₂ = 3 + 5 + 8 = 16 → |b₁| = √6, |b₂| = √50 = 5√2 → cos θ = 16/(√6 × 5√2) = 16/(10√3) = 8/(5√3) → θ = cos⁻¹(8/(5√3)).
Q9. Find the angle between the following pair of lines
(i) (x − 2)/2 = (y − 1)/5 = (z + 3)/(−3) and (x + 2)/(−1) = (y − 4)/8 = (z − 5)/4
(ii) x/2 = y/2 = z/1 and (x − 5)/4 = (y − 2)/1 = (z − 3)/8
Ans. cos θ = |a₁a₂ + b₁b₂ + c₁c₂| / (√(a₁² + b₁² + c₁²) √(a₂² + b₂² + c₂²))
(i) Direction ratios 2, 5, −3 and −1, 8, 4 → sum of products = −2 + 40 − 12 = 26 → magnitudes √38 and √81 = 9 → cos θ = 26/(9√38) → θ = cos⁻¹(26/(9√38)).
(ii) Direction ratios 2, 2, 1 and 4, 1, 8 → sum of products = 8 + 2 + 8 = 18 → magnitudes 3 and 9 → cos θ = 18/27 = 2/3 → θ = cos⁻¹(2/3).
Find the values of p so that the lines (1 − x)/3 = (7y − 14)/(2p) = (z − 3)/2 and (7 − 7x)/(3p) = (y − 5)/1 = (6 − z)/5 are at right angles.
Answer
Write both in standard form:
Line 1: (x − 1)/(−3) = (y − 2)/(2p/7) = (z − 3)/2
⇒ direction ratios −3, 2p/7, 2.
Line 2: (x − 1)/(−3p/7) = (y − 5)/1 = (z − 6)/(−5)
⇒ direction ratios −3p/7, 1, −5.
Perpendicular: (−3)(−3p/7) + (2p/7)(1) + 2(−5) = 0
⇒ 9p/7 + 2p/7 = 10
⇒ 11p/7 = 10
⇒ p = 70/11.
Show that the lines (x − 5)/7 = (y + 2)/(−5) = z/1 and x/1 = y/2 = z/3 are perpendicular to each other.
Answer
Direction ratios are 7, −5, 1 and 1, 2, 3.
7(1) + (−5)(2) + 1(3) = 7 − 10 + 3 = 0. Hence the lines are perpendicular.
Find the shortest distance between the lines r = (î + 2ĵ + k̂) + λ(î − ĵ + k̂) and r = 2î − ĵ − k̂ + μ(2î + ĵ + 2k̂).
Answer
a₁ = î + 2ĵ + k̂, b₁ = î − ĵ + k̂, a₂ = 2î − ĵ − k̂, b₂ = 2î + ĵ + 2k̂.
a₂ − a₁ = î − 3ĵ − 2k̂
b₁ × b₂ =
î | ĵ | k̂ |
1 | −1 | 1 |
2 | 1 | 2 |
= î(−2 − 1) − ĵ(2 − 2) + k̂(1 + 2) = −3î + 3k̂
⇒ |b₁ × b₂| = √18 = 3√2
(b₁ × b₂)·(a₂ − a₁) = −3(1) + 0 + 3(−2) = −9
d = |−9|/(3√2) = 3/√2 = 3√2/2 units.
Find the shortest distance between the lines (x + 1)/7 = (y + 1)/(−6) = (z + 1)/1 and (x − 3)/1 = (y − 5)/(−2) = (z − 7)/1.
Answer
a₁ = −î − ĵ − k̂, b₁ = 7î − 6ĵ + k̂; a₂ = 3î + 5ĵ + 7k̂, b₂ = î − 2ĵ + k̂.
a₂ − a₁ = 4î + 6ĵ + 8k̂
b₁ × b₂ =
î | ĵ | k̂ |
7 | −6 | 1 |
1 | −2 | 1 |
= î(−6 + 2) − ĵ(7 − 1) + k̂(−14 + 6) = −4î − 6ĵ − 8k̂
⇒ |b₁ × b₂| = √116 = 2√29
(b₁ × b₂)·(a₂ − a₁) = −16 − 36 − 64 = −116
d = 116/(2√29) = 58/√29 = 2√29 units.
Find the shortest distance between the lines whose vector equations are r = (î + 2ĵ + 3k̂) + λ(î − 3ĵ + 2k̂) and r = 4î + 5ĵ + 6k̂ + μ(2î + 3ĵ + k̂).
Answer
a₂ − a₁ = 3î + 3ĵ + 3k̂
b₁ × b₂ =
î | ĵ | k̂ |
1 | −3 | 2 |
2 | 3 | 1 |
= î(−3 − 6) − ĵ(1 − 4) + k̂(3 + 6) = −9î + 3ĵ + 9k̂
⇒ |b₁ × b₂| = √171 = 3√19
(b₁ × b₂)·(a₂ − a₁) = −27 + 9 + 27 = 9
d = 9/(3√19) = 3/√19 units.
Find the shortest distance between the lines whose vector equations are r = (1 − t)î + (t − 2)ĵ + (3 − 2t)k̂ and r = (s + 1)î + (2s − 1)ĵ − (2s + 1)k̂.
Answer
Rewrite: r = (î − 2ĵ + 3k̂) + t(−î + ĵ − 2k̂) and r = (î − ĵ − k̂) + s(î + 2ĵ − 2k̂).
a₂ − a₁ = 0î + ĵ − 4k̂
b₁ × b₂ =
î | ĵ | k̂ |
−1 | 1 | −2 |
1 | 2 | −2 |
= î(−2 + 4) − ĵ(2 + 2) + k̂(−2 − 1) = 2î − 4ĵ − 3k̂
⇒ |b₁ × b₂| = √29
(b₁ × b₂)·(a₂ − a₁) = 0 − 4 + 12 = 8
d = 8/√29 units.
Miscellaneous Exercise on Chapter 11
Find the angle between the lines whose direction ratios are a, b, c and b − c, c − a, a − b.
Answer
a(b − c) + b(c − a) + c(a − b) = ab − ac + bc − ab + ac − bc = 0
So cos θ = 0 → θ = 90°.
Find the equation of a line parallel to x-axis and passing through the origin.
Answer
The line passes through (0, 0, 0) with direction cosines 1, 0, 0.
Cartesian: x/1 = y/0 = z/0 (i.e., y = 0, z = 0). Vector: r = λî.
If the lines (x − 1)/(−3) = (y − 2)/(2k) = (z − 3)/2 and (x − 1)/(3k) = (y − 1)/1 = (z − 6)/(−5) are perpendicular, find the value of k.
Answer
Direction ratios −3, 2k, 2 and 3k, 1, −5.
(−3)(3k) + (2k)(1) + 2(−5) = 0
⇒ −9k + 2k − 10 = 0
⇒ −7k = 10
⇒ k = −10/7.
Find the shortest distance between lines r = 6î + 2ĵ + 2k̂ + λ(î − 2ĵ + 2k̂) and r = −4î − k̂ + μ(3î − 2ĵ − 2k̂).
Answer
a₂ − a₁ = −10î − 2ĵ − 3k̂
b₁ × b₂ =
î | ĵ | k̂ |
1 | −2 | 2 |
3 | −2 | −2 |
= î(4 + 4) − ĵ(−2 − 6) + k̂(−2 + 6) = 8î + 8ĵ + 4k̂
⇒ |b₁ × b₂| = √144 = 12
(b₁ × b₂)·(a₂ − a₁) = −80 − 16 − 12 = −108
d = 108/12 = 9 units.
Find the vector equation of the line passing through the point (1, 2, −4) and perpendicular to the two lines (x − 8)/3 = (y + 19)/(−16) = (z − 10)/7 and (x − 15)/3 = (y − 29)/8 = (z − 5)/(−5).
Answer
The required direction is perpendicular to both b₁ = 3î − 16ĵ + 7k̂ and b₂ = 3î + 8ĵ − 5k̂, so it is along b₁ × b₂:
î | ĵ | k̂ |
3 | −16 | 7 |
3 | 8 | −5 |
= î(80 − 56) − ĵ(−15 − 21) + k̂(24 + 48) = 24î + 36ĵ + 72k̂ = 12(2î + 3ĵ + 6k̂)
Required line: r = (î + 2ĵ − 4k̂) + λ(2î + 3ĵ + 6k̂).
Important Extra Questions
If (1/c, 1/c, 1/c) are the direction cosines of a line, find c.
Answer
3/c² = 1 → c² = 3 → c = ±√3.
Find the direction cosines of the line through the points (−2, 4, −5) and (1, 2, 3).
Answer
Direction ratios 3, −2, 8
⇒ √(9 + 4 + 64) = √77
⇒ direction cosines 3/√77, −2/√77, 8/√77.
Find the angle between the lines x/1 = y/2 = z/2 and x/2 = y/2 = z/1.
Answer
cos θ = (2 + 4 + 2)/(3 × 3) = 8/9
⇒ θ = cos⁻¹(8/9).
Show that the lines (x − 1)/2 = (y − 2)/3 = (z − 3)/4 and (x − 4)/5 = (y − 1)/2 = z/1 intersect.
Answer
a₂ − a₁ = 3î − ĵ − 3k̂; b₁ × b₂ = (2î + 3ĵ + 4k̂) × (5î + 2ĵ + k̂) = −5î + 18ĵ − 11k̂.
(b₁ × b₂)·(a₂ − a₁) = −15 − 18 + 33 = 0
⇒ shortest distance is 0, and the lines are not parallel, so they intersect.
Find the vector and cartesian equations of the line through (3, −2, −5) and (3, −2, 6).
Answer
Direction b = (3 − 3)î + (−2 + 2)ĵ + (6 + 5)k̂ = 11k̂.
Vector: r = (3î − 2ĵ − 5k̂) + λ(11k̂). Cartesian: (x − 3)/0 = (y + 2)/0 = (z + 5)/11, i.e., x = 3, y = −2.